【题目】 如图，在 △ A B C \triangle ABC △ABC 中，设 A B → = a → \overrightarrow{AB}=\overrightarrow{a} AB =a ， A C → = b → \overrightarrow{AC}=\overrightarrow{b} AC =b ， A P AP AP 的中点为 Q Q Q， B Q BQ BQ 的中点为 R R R， C R CR CR 的中点恰为 P P P，则 A P → = ( ) \overrightarrow{AP}=(\quad) AP =()
A . 1 2 a → + 1 2 b → A.\dfrac{1}{2}\overrightarrow{a}+\dfrac{1}{2}\overrightarrow{b} A.21a +21b
B . 1 3 a → + 2 3 b → B.\dfrac{1}{3}\overrightarrow{a}+\dfrac{2}{3}\overrightarrow{b} B.31a +32b
C . 2 7 a → + 4 7 b → C.\dfrac{2}{7}\overrightarrow{a}+\dfrac{4}{7}\overrightarrow{b} C.72a +74b
D . 4 7 a → + 2 7 b → D.\dfrac{4}{7}\overrightarrow{a}+\dfrac{2}{7}\overrightarrow{b} D.74a +72b
【解析】 因为 A P → + P C → = A P → + R P → = b → ⋯ ⋯ ① \overrightarrow{AP}+\overrightarrow{PC}=\overrightarrow{AP}+\overrightarrow{RP}=\overrightarrow{b}\qquad\cdots\cdots① AP +PC =AP +RP =b ⋯⋯① A Q → + Q B → = 1 2 A P → + 2 Q R → = a → ⋯ ⋯ ② \overrightarrow{AQ}+\overrightarrow{QB}=\dfrac{1}{2}\overrightarrow{AP}+2\overrightarrow{QR}=\overrightarrow{a}\quad\cdots\cdots② AQ +QB =21AP +2QR =a ⋯⋯② ① × 2 + ② ①\times2+② ①×2+②得： 5 2 A P → + 2 Q P → = 7 2 A P → = 2 b → + a → \dfrac{5}{2}\overrightarrow{AP}+2\overrightarrow{QP}=\dfrac{7}{2}\overrightarrow{AP}=2\overrightarrow{b}+\overrightarrow{a} 25AP +2QP =27AP =2b +a 即： A P → = 4 7 b → + 2 7 a → . \overrightarrow{AP}=\dfrac{4}{7}\overrightarrow{b}+\dfrac{2}{7}\overrightarrow{a}. AP =74b +72a . 即答案选 C.
